SOLUTION: In a survey of 628 males ages​ 18-64, 393 say they have gone to the dentist in the past year. Construct​ 90% and​ 95% confidence intervals for the population pro

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Question 1044004: In a survey of 628 males ages​ 18-64, 393 say they have gone to the dentist in the past year.
Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.
The​ 90% confidence interval for the population proportion p is? Round to three decimal places as​ needed.)
The​ 95% confidence interval for the population proportion p is? Round to three decimal places as​ needed.)
Interpret your results of both confidence intervals.
A.With the given​ confidence, it can be said that the sample proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.
B.With the given​ confidence, it can be said that the population proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.
C.With the given​ confidence, it can be said that the population proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is not between the endpoints of the given confidence interval.
Which interval is​ wider?
a)The​ 90% confidence interval
b)The​ 95% confidence interval
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
proportion(p) = 393 / 628 = 0.6258
:
standard error(se) = square root( 0.6258 * (1-0.6258) / 628 ) = 0.0193
:
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90% confidence interval(CI)
:
alpha(a) = 1 - (90/100) = 0.10
critical probability(p*) = 1 - (0.10/2) = 0.95
degrees of freedom(df) = 628 -1 = 627
critical value(use t statistic calculator) = 1.647
margin of error(me) = 1.647 * 0.0193 = 0.0318
90% ci is 0.626 + or - 0.032, that is, (0.594, 0.658)
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:
95% ci
:
a = 1 - (95/100) = 0.05
p* = 1 - (0.05/2) = 0.975
df = 627
cv = 1.964
me = 1.964 * 0.0193 = 0.0379
95% ci is 0.626 + or - 0.038, that is, (0.588, 0.664)
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:
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90% ci is (0.594, 0.658)
95% ci is (0.588, 0.664)
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:
We are trying to estimate the population proportion, answer is B
95% ci is wider, answer is b
:
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